$f(t) = -7t-3-h(t)$ $h(x) = -3x^{2}-x$ $g(t) = 7t^{2}-2t-3(h(t))$ $ g(f(0)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = (-7)(0)-3-h(0)$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = -3(0^{2})-0$ $h(0) = 0$ That means $f(0) = (-7)(0)-3-0$ $f(0) = -3$ Now we know that $f(0) = -3$ . Let's solve for $g(f(0))$ , which is $g(-3)$ $g(-3) = 7(-3)^{2}+(-2)(-3)-3(h(-3))$ To solve for the value of $g$ , we need to solve for the value of $h(-3)$ $h(-3) = -3(-3)^{2}-(-3)$ $h(-3) = -24$ That means $g(-3) = 7(-3)^{2}+(-2)(-3)+(-3)(-24)$ $g(-3) = 141$